If $G$ is abelian, then the set of all $g \in G$ such that $g = g^{-1}$ is a subgroup of $G$ 
Prove that if $G$ is abelian then the set $H$ of all elements of $G$ that are their own inverses is a subgroup of $G$.

Naturally in an abelian group, $ab = ba$ for $a, b \in G$, however I'm not sure how to show the set elements that are their own inverses is a subgroup of $G$ using arbitrary elements.
 A: $\newcommand{\N}{\Bbb N}$
Let $G$ an abelian group, let $e$ denote its identity element. For each $m\in\N$ define $$G(m):=\{g\in G: g^m=e\}.$$
$G(m)$ is a subgroup of $G$. Indeed, you can see that $e\in G(m)$. If $g,h\in G$, since $G$ is abelian we have
$$(gh^{-1})^m=g^m(h^{-1})^m=e(h^m)^{-1}=e^{-1}=e.$$
Therefore $G(m)\leq G$ as claimed.
A: A different way to phrase the same argument everyone gave:
The map $a\in G\mapsto a^2\in G$ is a group homomorphism and your subset $H$ is its kernel: it is therefore a subgroup of $G$.
A: You'll need to show only closure under multiplication (that is, that $ab\in H$ for all $a,b\in H$), since the identity is trivially its own inverse, so is in $H$, and since every element of $H$ is its own inverse, you don't need to check inverses, either. The fact that $G$ (so also $H$) is abelian makes checking closure fairly trivial.
A: Generally the one-step subgroup test is faster but in this case you can just check the group axioms: the only non-trivial one is closure. If $a^2=b^2=e$, can you see that $ab$ is its own inverse, given the group is Abelian?
A: Write H = {x in G: x*x = e}, where e is the identity element. 


*

*Show e is in H: since e*e = e, e is in H and so H is nonempty. 

*Consider x in H. Then x*x=e. Since the inverse of x is x and x is in H, H is closed under inverses. 

*Now consider x,y in H. Then x*x = e and y*y = e. So x*x*y*y = e. Since G is Abelian, so is H since H contains elements from G. So x*x*y*y = x*y*x*y = (x*y)*(x*y) = e, x*y is in H. So H is closed under multiplication. 
Thus H is a subgroup of G.  
